1. In the given circuit, with steady current, the potential drop across the capacitor must be.

PLease find the solution to the problem.

Apply Kirchoff law Take loop afdca2V-2RI-RI-V=0Now see .. V across AF =V across BE=V across CDSo. V(be)...RI+V= 2V-2RI (just reaarange the kirchoiff equation.)V(be)= 2v- 2R * v/3R = 4v/3Now Vbe= v acriss C + V So V(C)= v(be)-v= 4v/3-v= V/ 3... This is answer V/3

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